Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mn = 2e5 + 10;
const long long mod = 1e9 + 7;
long long l[mn], r[mn], len[mn];
int n;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  double y, xl, xr;
  cin >> y >> xl >> xr, y = -y;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
  len[0] = r[0] - l[0];
  for (int i = 1; i < n; i++) len[i] = len[i - 1] + r[i] - l[i];
  int q;
  cin >> q;
  while (q--) {
    double a, b;
    cin >> a >> b;
    double k = b / (b + y), c = a * y / (b + y);
    double lv = k * xl + c, rv = k * xr + c;
    int lef = upper_bound(l, l + n, (long long)floor(lv)) - l - 1;
    int rig = upper_bound(l, l + n, (long long)floor(rv)) - l - 1;
    double ans;
    if (lef == rig) {
      if (lef == -1) {
        ans = 0;
      } else {
        lv = min(lv, (double)r[lef]), rv = min(rv, (double)r[lef]);
        ans = rv - lv;
      }
    } else {
      ans = (rig - 1 >= 0 ? len[rig - 1] : 0) - (lef >= 0 ? len[lef] : 0);
      if (lef >= 0) lv = min(lv, (double)r[lef]);
      if (rig >= 0) rv = min(rv, (double)r[rig]);
      if (lef >= 0) ans += r[lef] - lv;
      if (rig >= 0) ans += rv - l[rig];
    }
    ans /= k;
    printf("%.6lf\n", ans);
  }
}